1. Field of the Invention
The present invention relates to integrated circuits, and, in particular, to transconductance amplifiers.
2. Description of the Related Art
A transconductance amplifier converts an input voltage signal into an output current signal, where output current is proportional to the input voltage. Conventional transconductance amplifiers have a transistor to provide amplification of the input signal and a DC bias circuit to bias the transistor for operations. In order to save power, e.g., in many battery-operated devices, circuits having transconductance amplifiers may be required to operate from low supply voltages. Typically, a certain minimum voltage is required in order to bias a transconductance amplifier for proper operation. As supply voltages decrease, the available voltage approaches, or in some cases drops below, the minimum voltage needed to bias the circuit.
FIG. 1 shows a traditional Gilbert cell mixer implementation that is used to illustrate this problem. In this circuit, three levels of transistors are stacked in the available power supply voltage. These are the current source formed by transistor B17, the transconductance stage which includes transistors B11 and B12, and the mixing core of transistors B13, B14, B15, and B16. Transistors B18 and B19 are part of a current mirror used to set the current in transistor B17. There is additional voltage drop across the current source resistor R15, the degeneration resistors R11 and R12, and the load resistors R13 and R14. At low power supply voltages required by some applications, it may become impossible to bias the transconductance stage for proper circuit operations.
FIG. 2 shows a modification to the traditional Gilbert cell mixer of FIG. 1 that helps to alleviate some of the biasing problems for a high-frequency, narrow-band circuit. Instead of using a separate current source such as was done with transistor B17 of FIG. 1, in the circuit of FIG. 2, transistors B21 and B22 perform double duty: providing both DC bias and AC gain. They, along with resistor R25, form a current mirror with transistors B28 and B29 and resistor R26 to set the proper DC bias level.
In order to eliminate the DC voltage drop and to reduce thermal noise, inductors L21 and L22 are used as degeneration devices instead of the resistors R11 and R12 of FIG. 1. At high frequencies, inductors L21 and L22 provide the necessary impedance for degeneration but do not suffer from DC voltage drop and do not contribute significant thermal noise.
Capacitors C21 and C22 allow the AC input signal to be coupled into the transconductance amplifier while simultaneously blocking any DC signals that could disrupt the bias scheme. Resistors R21 and R22 are used to provide a high differential impedance between the inputs to the transconductance amplifier. Because some base current flows through resistors R21 and R22 and creates a voltage drop, resistor R27 is used to create a compensating voltage drop in the path to the base of transistor B28. Inductor L23 provides an impedance at high frequencies that improves the common-mode performance of the mixer.
A problem exists with the circuit in FIG. 2 that can be explained by comparing the bias circuit in FIG. 1 to the bias circuit of FIG. 2. In FIG. 1, suppose that it is desired to operate the circuit such that the output of the current source (i.e., transistor B17) is M times larger than some current reference IREF1. Ignoring second-order effects, this may be accomplished by setting the area of transistor B17 to be M times larger than the area of transistor B18, and setting the value of resistor R16 to be M times the value of resistor R15. By scaling the transistor sizes, the same current density is maintained in transistors B18 and B17 resulting in an identical emitter-base voltage (Vbe) for both transistors. Let VREF1 be the voltage at node IF (this is approximately equal to IREF1 *R16), then the voltage at node 1E will be VREF1+Vbe -Vbe =VREF1. The current in resistor R15 is approximately equal to the output current of the current source and is equal to VREF1/R15=M*VREF1/R16=M*IREF1 as desired.
Similar DC operation is intended for the circuit of FIG. 2. In this circuit, it is assumed that the voltage drop across the inductors is zero and that half of the current flows through transistor B21 and half through transistor B22, with the total current flowing through resistor R25. If the total current in resistor R25 is desired to be M times IREF2, then transistors B21 and B22 are designed to be M/2 times as large as transistor B28. The value of resistor R26 is set to be M times larger than the value of resistor R25. The value of resistor R27 is set to be M/2 times as large as the values of resistor R21 or resistor R22, which are assumed to be equal in value. Under these conditions, the current in resistor R25 is M*IREF2, as desired.
An implicit assumption in the above analysis is that the resistance of the metal used for interconnection is zero. In reality, this is never the case, but it is usually low enough that it does not introduce a significant error. However, this is not necessarily a valid assumption for the circuit shown in FIG. 2. Integrated inductors are made of spirals of metal that can have significant resistance. If the resistance of the inductors is large enough compared to the value of resistor R25, significant errors will occur in the bias scheme. This is particularly true for large bias currents where resistor R25 will have a relatively low value. This could be addressed by adjusting the value of resistor R25 or resistor R26, but since the temperature characteristics of the resistors will not be the same as those of the metal, such a correction will introduce bias errors as temperature changes.
This problem is illustrated in FIG. 3, where the series resistance of the inductors is represented as resistors RMET31, RMET32, and RMET33. In order to compensate for this error, it is possible to add metal resistance in series with resistor R36 as represented by resistor RMET34 in FIG. 3. The value of resistor RMET34 needed to compensate for the inductor resistance(assuming resistor RMET31 is equal to resistor RMET32) is M * (RMET33+RMET31/2). The area needed to implement such a resistor can be very large and can add significant cost to an integrated circuit.